hello iam trying to solve a system of non linear equations and i don't know what the wrong so if any one can help me it will be appreciated..

function f= myfunction(z)
c2=z(1);
c3=z(2);
c4=z(3);
c5=z(4);
c6=z(5);
f(1)= 9*cos(c2) + 14.2*cos(c3) - 14.5*cos(c4)==17.7 ;
f(2)= 9*sin(c2) + 14.2*sin(c3) - 14.5*sin(c4) ;
f(3)= 9*sin(c2) + 12*sin(c5) == 9.3 ;
f(4)= 9*cos(c2) - 12*cos(c5) + sqrt(9.3^2 / (1-(cos(c6))^2));
end

In the Command window I pass in c2 as an initial value :

c2=0:0.01:pi;
fsolve(@myfunction,c2)

It gives me this error

Error using fsolve (line 281)

FSOLVE requires all values returned by functions to be of data type double.

Answer 1

[Image Analyst]

What is the class of f?

Since you're using "==" when you define it, I'm guessing it's not a double. Why are you using == anyway when you define f?

And why are you passing in 315 values for c2 (which becomes z inside the function) when you are only extracting and using the first 5 elements of the total 315 elements in z?

[Ahmed ElNahas]

c2 is my input angle and i want to know the values of c3,c4,c5,c6 as c2 changes.

so i put c2 in an interval from 0 to 360 degrees.

i removed '==' and still does not get the answer for the others ceta's

thanks in advance ..

[Image Analyst]

I have no idea what you want to do, even after your explanation. I don't even have an fsolve function. The best I can figure is that you want to call myfunction with a set of 5 elements. So, like this:

% Main Script
% Create data.
c2input = linspace(0, pi, 360);
% Call myfunction() for groups of 5 elements from c2.
for k = 1 : length(c2input)-4
	% Specify z as a group of 5 elements from c2input.
	z = c2input(k:k+4);
	% Compute f and pass in z.
	thisf = myfunction(z);
	% Print out values of f.
	fprintf('For k = %d, f = ', k);
	fprintf('%.3f ', thisf);
	fprintf('\n');
end

% This function is lower down in the same script m-file.
function f= myfunction(z)
c2=z(1);
c3=z(2);
c4=z(3);
c5=z(4);
c6=z(5);
f(1)= 9*cos(c2) + 14.2*cos(c3) - 14.5*cos(c4);
f(2)= 9*sin(c2) + 14.2*sin(c3) - 14.5*sin(c4);
f(3)= 9*sin(c2) + 12*sin(c5);
f(4)= 9*cos(c2) - 12*cos(c5) + sqrt(9.3^2 / (1-(cos(c6))^2));
end

[Ahmed ElNahas]

Sorry for misunderstanding

i have 4 equations with 4 unkonws (c3,c4,c5,c6) and c2 is my input so i want to solve this 4 non linear equations together to get the values of the unkowns thats all i want to do.

By knowing that c2 changes with time so that i write it in interval from 0 to pi .

Thats All and Thank you very much :)

[Image Analyst]

If equation 1 is supposed to give 17.7, and equation 3 is supposed to give 9.3, what are equations 2 and 4 supposed to give?

Answer 2

[Matt J]

It doesn't appear that there are exact solutions for c2 <= 1.26,

lb=zeros(1,4);
ub=2*pi*(lb+1);


c2=0:0.01:pi;
N=numel(c2);
resnorm=zeros(size(c2));
[z,ffinal]=deal(nan(4,N));

for i=N:-1:1;
    
 ci=c2(i);
 fun=@(z)myfunction(z,ci);
 
 if i<N && norm(ffinal(:,i+1))<1e-4
   z0=z(:,i+1);
 else
  [z0,resnorm(i)]=ga(@(z) norm(fun(z)),4,[],[],[],[],lb,ub); %global sweep

 end
 
 [z(:,i),ffinal(:,i)]=fsolve(fun,z0); %refine

end

plot(c2.',ffinal.','x'); legend('f1','f2','f3','f4')

ylabel 'Final f(i)'
xlabel 'c2'


function f= myfunction(z,c2)

    c3=z(1);
    c4=z(2);
    c5=z(3);
    c6=z(4);
    f(1)= 9*cos(c2) + 14.2*cos(c3) - 14.5*cos(c4)-17.7 ;
    f(2)= 9*sin(c2) + 14.2*sin(c3) - 14.5*sin(c4) ;
    f(3)= 9*sin(c2) + 12*sin(c5) - 9.3 ;
    f(4)= 9*cos(c2) - 12*cos(c5) + sqrt(9.3^2 / (1-(cos(c6))^2));
end

[Image Analyst]

Ahmed forgot to list his version number. My R2019b help does not list an fsolve() function. I tried to run it and it says it requires the Global Optimization Toolbox. I've added that to the Products list.

[Matt J]

@Image Analyst.

fsolve is in the Optimization Toolbox and ga is in the Global Optimization Toolbox,

@Ahmed,

If I change the final equation to

f(4)= (9*cos(c2) - 12*cos(c5)).^2 - (9.3^2 / (1-(cos(c6))^2));

then solutions are found over a wider range of c2.